Periodic Table JEE Main PYQs With Solutions and Most Important MCQs pdf download

Anand Classes provides the Periodic Table JEE Main PYQs With Solutions and Most Important MCQs pdf download, specially designed for students preparing for JEE Main and advanced exams. This comprehensive study material includes previous years’ questions, detailed solutions, and practice questions that are crucial for understanding patterns and trends in the periodic table. Perfect for self-study or classroom learning, these notes help you grasp essential concepts, improve problem-solving speed, and enhance accuracy. Click the print button to download study material and notes.

JEE Main 2021 Question:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Metallic character decreases and non-metallic character increases on moving from left to right in a period.
Reason (R): It is due to increase in ionisation enthalpy and decrease in electron gain enthalpy, when one moves from left to right in a period.
(A) (A) is false but (R) is true.
(B) (A) is true but (R) is false.
(C) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(D) Both (A) and (R) are correct but (R) is not the correct explanation of (A).

Answer — Correct Option: (B) (A) is true but (R) is false

Step 1: Concept explanation / rule / definition
Across a period (left → right) effective nuclear charge ($Z_{\text{eff}}$) on valence electrons increases. As $Z_{\text{eff}}$ increases, atoms hold their electrons more tightly: it becomes harder to lose electrons (metallic character decreases) and easier to gain electrons (non-metallic character increases). Ionisation enthalpy ($\mathrm{IE}$) therefore increases across a period. Electron gain enthalpy (EGH) generally becomes more negative (more exothermic) for non-metals across a period, indicating a greater tendency to gain electrons.

Step 2: Identify elements or given data

• Assertion (A): Trend of metallic and non-metallic character across a period (decrease/increase respectively).
• Reason (R): Causal statement attributing the trend to “increase in ionisation enthalpy and decrease in electron gain enthalpy” across the period.

Step 3: Comparison / solving process

1. The first part of R (“increase in ionisation enthalpy”) is correct and does explain why metallic character decreases.
2. The second part of R (“decrease in electron gain enthalpy“) is ambiguous/misleading: electron gain enthalpy values become more negative (i.e., greater magnitude of energy released) across a period for many elements — this means the tendency to gain electrons increases, not decreases. Saying “decrease in electron gain enthalpy” can be interpreted as the algebraic value becoming less negative (which is incorrect). Because R contains this incorrect/misleading statement, R as given is false even though it contains a correct partial idea.
3. Since A is correctly stated but R is not correctly stated as a reason (due to the wrong phrasing about electron gain enthalpy), the correct choice is that A is true and R is false.

Final Answer
$$
\boxed{(B)\ \text{(A) is true but (R) is false}}
$$

Concept Takeaway

• Exam tip: Across a period, remember $Z_{text{eff}}$ ↑ ⇒ IE ↑ ⇒ metallic character ↓. Also remember that electron gain enthalpy becomes more negative for many non-metals (greater tendency to gain electrons).
• Concept explained simply: Increased nuclear attraction across a period makes atoms less willing to lose electrons and more willing to gain them. R is partly right (IE ↑) but incorrectly worded about electron gain enthalpy.
• download notes, study material, preparation notes, class 11 chemistry, JEE PYQs chapterwise, Anand Classes, Periodic Trends

JEE Main 2021 Question:
The correct order of ionic radii for the ions P³⁻, S²⁻, Ca²⁺, K⁺, Cl⁻ is:
(A) P³⁻ > S²⁻ > Cl⁻ > K⁺ > Ca²⁺
(B) Cl⁻ > S²⁻ > P³⁻ > Ca²⁺ > K⁺
(C) P³⁻ > S²⁻ > Cl⁻ > Ca²⁺ > K⁺
(D) K⁺ > Ca²⁺ > P³⁻ > S²⁻ > Cl⁻

Answer — Correct Option: (A) P³⁻ > S²⁻ > Cl⁻ > K⁺ > Ca²⁺

Step 1: Concept explanation / rule
For isoelectronic species (same number of electrons), ionic radius decreases as the nuclear charge (number of protons) increases. Greater nuclear charge pulls the same number of electrons more strongly, reducing size.

Step 2: Identify given ions and electron count

• $P^{3-}$ : Atomic number 15 → Electrons = $15+3=18$
• $S^{2-}$ : Atomic number 16 → Electrons = $16+2=18$
• $Cl^{-}$ : Atomic number 17 → Electrons = $17+1=18$
• $K^{+}$ : Atomic number 19 → Electrons = $19-1=18$
• $Ca^{2+}$ : Atomic number 20 → Electrons = $20-2=18$

All have 18 electrons → they are isoelectronic.

Step 3: Compare by nuclear charge

• $P^{3-}$ → 15 protons (least nuclear pull → largest radius)
• $S^{2-}$ → 16 protons
• $Cl^{-}$ → 17 protons
• $K^{+}$ → 19 protons
• $Ca^{2+}$ → 20 protons (strongest nuclear pull → smallest radius)

Thus, order of ionic radii is:
$$ P^{3-} > S^{2-} > Cl^{-} > K^{+} > Ca^{2+} $$

Final Answer
$$ \boxed{(A)\ P^{3-} > S^{2-} > Cl^{-} > K^{+} > Ca^{2+}} $$

Concept Takeaway

• Exam tip: For isoelectronic ions, always compare proton count (Z) — higher Z ⇒ smaller radius.
• Remember: More negative charge ⇒ larger ion, more positive charge ⇒ smaller ion.
• download notes, study material, preparation notes, class 11 chemistry periodic trends, JEE PYQs chapterwise, Anand Classes, Periodic Properties

JEE Main 2021 Question:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Barium carbonate is insoluble in water and is highly stable.
Reason (R): The thermal stability of the carbonates increases with increasing cationic size.
(A) Both (A) and (R) are true and (R) is the true explanation of (A)
(B) (A) is true but (R) is false
(C) Both (A) and (R) are true but (R) is not the true explanation of (A)
(D) (A) is false but (R) is true.

Answer — Correct Option: (A) Both (A) and (R) are true and (R) is the true explanation of (A)

Step 1: Concept explanation / rule / definition
Thermal stability of group 2 (alkaline earth) carbonates increases down the group because larger cations have lower polarizing power. Lower polarization means less distortion of the carbonate ion ($\mathrm{CO_3^{2-}}$), resulting in stronger ionic character and higher thermal stability. Solubility of carbonates of group 2 metals generally decreases down the group; heavier carbonates like barium carbonate ($\mathrm{BaCO_3}$) are sparingly soluble (commonly treated as insoluble) and thermally more stable than those of lighter cations.

Step 2: Identify elements or given data

• Assertion: $\mathrm{BaCO_3}$ is insoluble in water and highly stable.
• Reason: Thermal stability of carbonates increases with increasing cationic size (down the group).

Step 3: Comparison / solving process

1. Barium ($\mathrm{Ba^{2+}}$) is a large cation (atomic number 56). Its large size gives it low polarizing power compared to smaller cations (e.g., $\mathrm{Mg^{2+}}$, $\mathrm{Ca^{2+}}$).
2. Low polarizing power means the $\mathrm{CO_3^{2-}}$ anion remains less distorted, bonds remain more ionic, and decomposition on heating is more difficult → higher thermal stability.
3. Solubility: heavier alkaline earth carbonates are less soluble; $\mathrm{BaCO_3}$ is sparingly soluble and is treated as practically insoluble in water in standard qualitative contexts.
4. Thus Reason (R) correctly states the underlying principle that explains Assertion (A).

Final Answer
$$
\boxed{(A)\ \text{Both (A) and (R) are true and (R) is the true explanation of (A)}}
$$

Concept Takeaway

• Exam tip: For group 2 carbonates remember down the group → cation size ↑ → polarizing power ↓ → carbonate stability ↑; this explains why $\mathrm{BaCO_3}$ is more stable and less soluble than $\mathrm{CaCO_3}$ or $\mathrm{MgCO_3}$.
• Concept explained simply: larger cations distort the anion less, so heavy-metal carbonates resist thermal decomposition and dissolve less.
• Integrated resources note: practice periodic trends and group-wise stability with chapterwise JEE PYQs from Class 11 chemistry — this is useful for revision and for materials like Anand Classes study notes and preparation resources.

JEE Main 2021 Question:
The correct order of first ionisation enthalpy is:
(A) Mg < S < Al < P
(B) Mg < Al < S < P
(C) Al < Mg < S < P
(D) Mg < Al < P < S

Answer — Correct Option: (C) Al < Mg < S < P

Step 1: Concept explanation / rule / definition
Ionisation enthalpy is the energy required to remove the outermost electron from an isolated gaseous atom. Across a period, it generally increases due to increasing effective nuclear charge and decreasing atomic radius. However, exceptions occur due to electronic configuration stability (completely filled and half-filled subshells).

Step 2: Identify elements and electron configurations

• $Mg$ (Z = 12): $[Ne]\,3s^2$ → stable, filled $3s$ subshell.
• $Al$ (Z = 13): $[Ne]\,3s^2 3p^1$ → single $3p$ electron is relatively easy to remove.
• $S$ (Z = 16): $[Ne]\,3s^2 3p^4$ → removal from $p^4$ (more repulsion, slightly less stable).
• $P$ (Z = 15): $[Ne]\,3s^2 3p^3$ → half-filled $p^3$ subshell gives extra stability.

Step 3: Comparison / solving process

1. $Al$ vs. $Mg$: Al should have higher IE as it is to the right, but due to $3p^1$ electron being easier to remove, $IE(Al) < IE(Mg)$.
2. $S$ vs. $P$: $P$ has a half-filled $3p^3$ subshell (extra stable), so $IE(P) > IE(S)$.
3. Combining: Al < Mg < S < P.

Final Answer
$$
\boxed{(C)\ Al < Mg < S < P}
$$

Concept Takeaway

• Exam tip: Remember the two main exceptions across a period — $Mg > Al$ (due to stable $3s^2$ vs $3p^1$) and $P > S$ (due to half-filled $p^3$ stability).
• Explained simply: filled and half-filled subshells resist electron removal, creating deviations from the general trend.
• For effective JEE preparation, practice periodic trends and their exceptions using Class 11 Chemistry notes, JEE PYQs, and Anand Classes chapterwise material on Periodic Properties.

JEE Main 2021 Question:
The ionic radii of $F^-$ and $O^{2-}$ respectively are $1.33 \,\text{Å}$ and $1.40 \,\text{Å}$, while the covalent radius of $N$ is $0.74 \,\text{Å}$.
The correct statement for the ionic radius of $N^{3-}$ from the following is:
(A) It is smaller than $F^-$ and $N$
(B) It is bigger than $O^{2-}$ and $F^-$
(C) It is bigger than $F^-$ and $N$, but smaller than $O^{2-}$
(D) It is smaller than $O^{2-}$ and $F^-$, but bigger than $N$

Answer — Correct Option: (B) It is bigger than $O^{2-}$ and $F^-$

Step 1: Concept explanation
For isoelectronic species, the ionic radius decreases as the nuclear charge increases. More protons pull the same number of electrons closer to the nucleus, reducing the size.

Step 2: Identify given data

• $F^-$ radius = $1.33 \,\text{Å}$
• $O^{2-}$ radius = $1.40 \,\text{Å}$
• Covalent radius of $N$ = $0.74 \,\text{Å}$
• Ions compared: $N^{3-}$, $O^{2-}$, $F^-$
• All are isoelectronic ($10$ electrons each).

Step 3: Nuclear charges

• $N$: 7 protons
• $O$: 8 protons
• $F$: 9 protons

Lower nuclear charge → weaker pull on electrons → larger ionic radius.

So the order of radii is:
$$
N^{3-} > O^{2-} > F^-
$$

Step 4: Compare with covalent radius
The covalent radius of neutral $N$ ($0.74 \,\text{Å}$) is much smaller than ionic $N^{3-}$, because extra electrons increase repulsion and expand the size.

Step 5: Evaluate options

• (A) Smaller than $F^-$ and $N$: ❌ incorrect.
• (B) Bigger than $O^{2-}$ and $F^-$: ✅ correct.
• (C) Bigger than $F^-$ and $N$, but smaller than $O^{2-}$: ❌ incorrect.
• (D) Smaller than $O^{2-}$ and $F^-$, but bigger than $N$: ❌ incorrect.

Final Answer
$$
\boxed{(B)\ N^{3-} \text{ is bigger than } O^{2-} \text{ and } F^-}
$$

Concept Takeaway

• For isoelectronic species, greater nuclear charge = smaller size.
• $N^{3-}$ has the largest radius, while $F^-$ has the smallest.
• Useful for solving questions on periodic properties and trends of ionic radii in JEE Chemistry.
• Study with chapterwise JEE PYQs, Class 11 chemistry notes, ionic radii trends, and Anand Classes material for exam preparation.

JEE Main 2021 Question:
The ionic radii of $K^+$, $Na^+$, $Al^{3+}$ and $Mg^{2+}$ are in the order:
(A) $Na^+ < K^+ < Mg^{2+} < Al^{3+}$
(B) $Al^{3+} < Mg^{2+} < K^+ < Na^+$
(C) $Al^{3+} < Mg^{2+} < Na^+ < K^+$
(D) $K^+ < Al^{3+} < Mg^{2+} < Na^+$

Answer — Correct Option: (C) $Al^{3+} < Mg^{2+} < Na^+ < K^+$

Step 1: Concept explanation
Ionic radius depends on nuclear charge and electron configuration. For isoelectronic ions (same number of electrons), greater nuclear charge → smaller radius.

Step 2: Count electrons (check isoelectronic nature)

• $K^+$: $19 – 1 = 18$ electrons
• $Na^+$: $11 – 1 = 10$ electrons
• $Mg^{2+}$: $12 – 2 = 10$ electrons
• $Al^{3+}$: $13 – 3 = 10$ electrons

So, $Na^+$, $Mg^{2+}$, and $Al^{3+}$ are isoelectronic with 10 electrons (like Neon). $K^+$ has 18 electrons (like Argon), not isoelectronic with the others.

Step 3: Compare radii among isoelectronic ions

• $Al^{3+}$: highest nuclear charge (13 protons), smallest radius
• $Mg^{2+}$: 12 protons, radius larger than $Al^{3+}$
• $Na^+$: 11 protons, largest among the 10-electron ions

Order within the 10-electron group:
$$Al^{3+} < Mg^{2+} < Na^+$$

Step 4: Compare with $K^+$
$K^+$ has 18 electrons (argon configuration), so its radius is much larger than $Na^+$. $K^+$ is the largest ion among the given species.

Final Order of Ionic Radii
$$Al^{3+} < Mg^{2+} < Na^+ < K^+$$

Correct Answer:
(C) $Al^{3+} < Mg^{2+} < Na^+ < K^+$

Concept Takeaway

• For isoelectronic species: more protons = smaller size
• $K^+$ is an exception, being larger due to extra electron shells
• Important for JEE Chemistry periodic properties, ionic radius trends, and competitive exam preparation (Class 11 Chemistry, PYQs, Anand Classes notes)

JEE Main 2021 Question:
The incorrect statement about D.I. Mendeleev is:
(A) He authored the textbook – Principles of Chemistry
(B) At the time, he proposed Periodic Table of elements, structure of atom was known
(C) Element with atomic number 101 is named after him
(D) He invented accurate barometer

Answer — Correct Option: (B) At the time, he proposed Periodic Table of elements, structure of atom was known

Step 1: Concept explanation

• Dmitri Mendeleev published the first version of the periodic table in 1869.
• At that time, the atomic structure was not yet understood; concepts like electrons, protons, neutrons, and their arrangement were unknown.
• Mendeleev organized elements based on atomic weights and recurring chemical properties, not atomic numbers.
• The relationship between atomic number and number of protons was established later by Henry Moseley in the early 20th century.

Step 2: Evaluate other statements

• (A) ✅ Correct — Mendeleev authored Principles of Chemistry, a widely influential textbook.
• (C) ✅ Correct — Element 101, Mendelevium (Md), is named in his honor.
• (D) ⚠ Minor clarification — Mendeleev did not invent the barometer; that was Evangelista Torricelli in the 17th century.

Final Answer:
Structure of atom was not known when Mendeleev proposed the periodic table

Concept Takeaway

• Key JEE tip: Mendeleev’s periodic table was based on atomic weights and chemical properties, not atomic structure.
• Historical accuracy: atomic number concept came later (Moseley, 1913).
• Useful for Class 11 Chemistry, JEE PYQs, Anand Classes study material on periodic table history.

JEE Main 2021 Question:
The first ionization energy of magnesium is smaller as compared to that of elements X and Y, but higher than that of Z. The elements X, Y, and Z, respectively, are:
(A) neon, sodium and chlorine
(B) argon, chlorine and sodium
(C) chlorine, lithium and sodium
(D) argon, lithium and sodium

Answer — Correct Option: (B) argon, chlorine and sodium

Step 1: Analyze the given information

• First ionization energy (IE) of magnesium (Mg) satisfies:
  $$IE_{Mg} < IE_X \text{ and } IE_{Mg} < IE_Y$$ $$IE_{Mg} > IE_Z$$

Step 2: Recall trends in first ionization energy

• Across a period: IE increases (nuclear charge ↑, atomic radius ↓)
• Down a group: IE decreases (atomic radius ↑, shielding ↑)
• Noble gases: very high IE
• Alkali metals: very low IE

Step 3: Position of magnesium

• Mg (Z = 12) is an alkaline earth metal, Group 2, Period 3.

Step 4: Evaluate options

• Option A: neon, sodium, chlorine
• Ne (noble gas) → IE >> Mg ✅
• Na (alkali metal) → IE << Mg ❌
• Cl → IE > Mg ✅
• Doesn’t fit because Na has lower IE but is proposed as X or Y.
• Option B: argon, chlorine, sodium
• Ar → IE > Mg ✅
• Cl → IE > Mg ✅
• Na → IE < Mg ✅
• Fits all conditions: X = Ar, Y = Cl (higher IE), Z = Na (lower IE).
• Option C: chlorine, lithium, sodium
• Cl → IE > Mg ✅
• Li → IE < Mg ❌
• Na → IE < Mg ❌
• Doesn’t fit; only one element should have lower IE (Z).
• Option D: argon, lithium, sodium
• Ar → IE > Mg ✅
• Li → IE < Mg ❌
• Na → IE < Mg ❌
• Doesn’t fit; both Li and Na have lower IE than Mg.

Step 5: Confirm the correct option

• Option B satisfies all the given conditions relative to Mg.

Final Answer:
$$
\boxed{(B)\ \text{argon, chlorine and sodium}}
$$

Concept Takeaway

• Ionization energy trends: increases across a period, decreases down a group.
• Noble gases have highest IE; alkali metals have lowest IE.
• Useful for JEE Chemistry periodic properties, first ionization energy trends, Class 11 Chemistry, Anand Classes PYQs and chapterwise notes.

JEE Main 2021 Question:
The ionic radius of $Na^+$ ion is 1.02 Å. The ionic radii (in Å) of $Mg^{2+}$ and $Al^{3+}$, respectively, are:
(A) 1.05 and 0.99
(B) 0.72 and 0.54
(C) 0.85 and 0.99
(D) 0.68 and 0.72

Answer — Correct Option: (B) 0.72 and 0.54

Step 1: Understand the trend of ionic radius in isoelectronic species

• For isoelectronic species (same number of electrons), ionic radius decreases as nuclear charge increases.
• Higher nuclear charge pulls electrons closer, resulting in smaller ionic radius.

Step 2: Identify the isoelectronic species

• $Na^+$, $Mg^{2+}$, and $Al^{3+}$ all have 10 electrons (like Neon).

Step 3: Compare nuclear charges

• $Na$: 11 protons
• $Mg$: 12 protons
• $Al$: 13 protons

Step 4: Apply the trend to determine ionic radii

• Increasing nuclear charge from $Na^+ \to Mg^{2+} \to Al^{3+}$ → ionic radius decreases.
• Given $Na^+$ radius = 1.02 Å, radii of $Mg^{2+}$ and $Al^{3+}$ should be smaller than 1.02 Å and follow Mg²⁺ > Al³⁺.

Step 5: Evaluate the given options

• (A) 1.05 and 0.99 → Mg²⁺ larger than Na⁺ ❌
• (B) 0.72 and 0.54 → correct trend ✅
• (C) 0.85 and 0.99 → Al³⁺ larger than Mg²⁺ ❌
• (D) 0.68 and 0.72 → Al³⁺ larger than Mg²⁺ ❌

Final Answer:
$$
\boxed{(B)\ 0.72 \text{ Å and } 0.54 \text{ Å}}
$$

Concept Takeaway

• For isoelectronic ions, more protons → smaller ionic radius.
• This trend is critical for JEE Chemistry periodic table properties, ionic radius comparisons, and Class 11 Chemistry preparation notes (Anand Classes).

JEE main 2021 Question:
The characteristics of elements X, Y and Z with atomic numbers 33, 53, and 83, respectively, are:
(A) X and Y are metalloids and Z is a metal
(B) X and Z are non-metals and Y is a metalloid
(C) X is a metalloid, Y is a non-metal and Z is a metal
(D) X, Y and Z are metals

Answer — Correct Option: (C) X is a metalloid, Y is a non-metal and Z is a metal

Step 1: Locate elements on the periodic table

• Element X (Z = 33): Arsenic (As) → Located on the stair-step line, separates metals and non-metals → Metalloid
• Element Y (Z = 53): Iodine (I) → Group 17 (halogens) → Non-metal
• Element Z (Z = 83): Bismuth (Bi) → Group 15, post-transition metal → Metal

Step 2: Classify based on periodic trends

• Metalloids: semiconducting properties, staircase line elements (As, Sb, Te, etc.)
• Non-metals: high electronegativity, typically right side of periodic table (I, Cl, F, etc.)
• Metals: malleable, conductive, left and center of periodic table (Bi, Sn, Pb, etc.)

Final Answer:
$$
\boxed{(C)\ X \text{ is a metalloid, } Y \text{ is a non-metal, and } Z \text{ is a metal}}
$$

Concept Takeaway

• Periodic table positioning determines metallic, non-metallic, or metalloid character.
• Important for JEE Chemistry classification of elements, periodic trends, Class 11 Chemistry notes, and Anand Classes preparation material.

JEE Main 2021 Question:
Which pair of oxides is acidic in nature?
(A) $CaO, SiO_2$
(B) $B_2O_3, CaO$
(C) $B_2O_3, SiO_2$
(D) $N_2O, BaO$

Answer — Correct Option: (C) $B_2O_3, SiO_2$

Step 1: Understand acidic oxides

• Acidic oxides are usually formed by non-metals.
• They react with water to form acids or react with bases to form salt and water.
• Examples:
• $SiO_2$ reacts with strong bases like $NaOH$ to form silicates.
• $B_2O_3$ reacts with water or bases to form borates.

Step 2: Understand basic and neutral oxides

• Basic oxides are formed by metals, react with water to form bases.
• Examples: $CaO$, $BaO$
• Neutral oxides do not react with acids or bases.
• Example: $N_2O$

Step 3: Compare given options

• (A) $CaO$ is basic, $SiO_2$ is acidic ❌
• (B) $B_2O_3$ acidic, $CaO$ basic ❌
• (C) $B_2O_3$ acidic, $SiO_2$ acidic ✅
• (D) $N_2O$ neutral, $BaO$ basic ❌

Final Answer:
$$
\boxed{(C)\ B_2O_3, SiO_2}
$$

Concept Takeaway

• Non-metal oxides → acidic, metal oxides → basic, some oxides → neutral.
• Crucial for JEE Chemistry periodic trends, oxides reactions, Class 11 Chemistry, and Anand Classes preparation notes.

JEE Main 2021 Question:
The correct order of electron gain enthalpy for the elements O, S, Se, Te is:
(A) Te > Se > S > O
(B) S > Se > Te > O
(C) S > O > Se > Te
(D) O > S > Se > Te

Answer — Correct Option: (B) S > Se > Te > O

Step 1: Understand electron gain enthalpy (EGE)

• Electron gain enthalpy is the energy change when an atom gains an electron.
• More negative EGE → easier to gain an electron.

Step 2: Trend in a group

• Down a group: Atomic size increases → electron-electron repulsion increases → EGE becomes less negative.
• For group 16 elements: S > Se > Te (in terms of electron gain enthalpy magnitude).

Step 3: Anomaly with Oxygen

• Oxygen has small atomic size → high electron density → strong repulsion when an extra electron is added to 2p orbital.
• This makes O’s EGE less negative than S, Se, and Te.

Step 4: Combine trend and anomaly

• Considering the normal group trend and oxygen anomaly:
  $$
  \text{S} > \text{Se} > \text{Te} > \text{O}
  $$

Final Answer:
$$
\boxed{(B)\ S > Se > Te > O}
$$

Concept Takeaway

• Electron gain enthalpy becomes less negative down a group.
• Oxygen anomaly occurs due to small size and high electron-electron repulsion.
• Important for JEE Chemistry periodic properties, Group 16 elements, Class 11 Chemistry notes, Anand Classes preparation material.

JEE Main 2021 Question:
The correct order of bond dissociation enthalpy for halogens is:
(A) $F_2 > Cl_2 > Br_2 > I_2$
(B) $Cl_2 > Br_2 > F_2 > I_2$
(C) $Cl_2 > Br_2 > F_2 > I_2$
(D) $Cl_2 > F_2 > Br_2 > I_2$

Answer — Correct Option: (C) $Cl_2 > Br_2 > F_2 > I_2$

Step 1: Understand bond dissociation enthalpy (BDE)

• BDE is the energy required to break a bond homolytically.
• Trend down the group: Larger atomic size → longer bond length → weaker bond → lower BDE.

Step 2: Halogen BDE trend

• Expected trend: $F_2 > Cl_2 > Br_2 > I_2$ (based on atomic size).
• Fluorine anomaly: Small F atoms → strong lone pair-lone pair repulsion → weakens F-F bond → lower BDE than Cl₂ and Br₂.

Step 3: Correct order considering anomaly

• $Cl_2 > Br_2 > F_2 > I_2$

Final Answer:
$$
\boxed{Cl_2 > Br_2 > F_2 > I_2}
$$

Concept Takeaway

• Bond dissociation enthalpy decreases down a group due to increasing bond length.
• Fluorine anomaly occurs due to lone pair repulsion.
• Important for JEE Chemistry periodic trends, halogen chemistry, Class 11 Chemistry, and Anand Classes study material.

JEE Main 2021 Question:
Consider the elements Mg, Al, Si, S, and P. The correct increasing order of first ionization enthalpy is:
(A) Al < Mg < Si < S < P
(B) Al < Mg < S < Si < P
(C) Mg < Al < Si < S < P
(D) Mg < Al < Si < P < S

Answer — Correct Option: (A) Al < Mg < Si < S < P

Step 1: Recall ionization enthalpy (IE) trends

• IE increases across a period due to increasing nuclear charge and decreasing atomic radius.
• IE decreases down a group due to increasing atomic size and shielding.
• Exceptions occur due to electron configuration stability.

Step 2: Analyze the elements in Period 3

• Mg (3s²): Stable filled s-orbital → higher IE than Al?
• Al (3p¹): Single p-electron → easier to remove → lower IE than Mg
• Si (3p²): Two p-electrons → higher IE than Mg and Al
• P (3p³): Half-filled p-orbital → most stable → highest IE
• S (3p⁴): Paired p-electrons → repulsion → easier to remove → IE less than P

Step 3: Arrange in increasing order

• Considering exceptions due to electron configuration:
  $$
  \text{Al} < \text{Mg} < \text{Si} < \text{S} < \text{P}
  $$

Final Answer:
$$
\boxed{\text{Al} < \text{Mg} < \text{Si} < \text{S} < \text{P}}
$$

Concept Takeaway

• Ionization enthalpy increases across a period with exceptions caused by half-filled and fully filled orbitals.
• Crucial for JEE Chemistry periodic trends, Period 3 elements, Class 11 Chemistry preparation notes, Anand Classes study material.

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